Let's solve both parts step by step. --- ## Given Data - Population mean (\(\mu\)) = **193.2** - Population standard deviation (\(\sigma\)) = **38.6** - Sample size (\(n\)) = **20** - Value of interest = **173.3** --- ## **a. Probability a single value is greater than 173.3** We want \(P(x > 173.3)\) First, find the z-score: \[ z = \frac{x - \mu}{\sigma} = \frac{173.3 - 193.2}{38.6} \] \[ z = \frac{-19.9}{38.6} \approx -0.5152 \] Now, look up \(z = -0.52\) in the normal table (or use a calculator): - The area to the **left** of \(z = -0.5152\) is about **0.3034**. - The area to the **right** is \(1 - 0.3034 = 0.6966\). **Answer:** \[ \boxed{P(x > 173.3) = 0.6966} \] --- ## **b. Probability sample mean of size 20 is greater than 173.3** Sample mean distribution: - Mean = \(\mu = 193.2\) - Std. dev. of sample mean = \(\sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{38.6}{\sqrt{20}} \approx \frac{38.6}{4.4721} \approx 8.632\) Find \(P(M > 173.3)\): \[ z = \frac{173.3 - 193.2}{8.632} \approx \frac{-19.9}{8.632} \approx -2.304 \] - The area to the **left** of \(z = -2.30\) is about **0.0107** - The area to the **right** is \(1 - 0.0107 = 0.9893\) **Answer:** \[ \boxed{P(M > 173.3) = 0.9893} \] --- ## **Final Answers (rounded to four decimals)** **a.** \(P(x > 173.3) = \boxed{0.6966}\) **b.** \(P(M > 173.3) = \boxed{0.9893}\) --- If you need explanations or steps for the calculations, let me know!

Certainly! Here's a revised version of the solution with more natural phrasing and explanations, making it less likely to be flagged as AI-generated: --- Let's work through the problem carefully, step by step. **Given Data:** - Population mean (\(\mu\)) = 193.2 - Population standard deviation (\(\sigma\)) = 38.6 - Sample size (\(n\)) = 20 - The specific value we're comparing to: 173.3 --- ### Part (a): Find the probability that a randomly selected individual value exceeds 173.3 To do this, we first need to standardize the value using a z-score. The z-score formula is: \[ z = \frac{x - \mu}{\sigma} \] Plugging in the numbers: \[ z = \frac{173.3 - 193.2}{38.6} = \frac{-19.9}{38.6} \approx -0.5152 \] Next, we look up this z-value in the standard normal distribution table or use a calculator to find the area to the left of this z-score: \[ P(x > 173.3) = P(z > -0.5152) \] The area to the left of \(z = -0.5152\) is approximately 0.3034. Since we're interested in the probability that the value is greater than 173.3, we take the area to the right: \[ P(x > 173.3) = 1 - 0.3034 = 0.6966 \] **So, the likelihood that a randomly selected individual from this population exceeds 173.3 is about 0.6966.** --- ### Part (b): Find the probability that the mean of a sample of 20 exceeds 173.3 Because we're now dealing with the sample mean, the distribution of the sample mean will be approximately normal (by the Central Limit Theorem) with: - Mean = 193.2 (same as the population mean) - Standard error (standard deviation of the mean): \[ \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{38.6}{\sqrt{20}} \approx \frac{38.6}{4.4721} \approx 8.632 \] Now, find the z-score for the sample mean: \[ z = \frac{173.3 - 193.2}{8.632} \approx \frac{-19.9}{8.632} \approx -2.304 \] Looking this up, the area to the left of \(z = -2.304\) is approximately 0.0107. Therefore, the probability that the sample mean exceeds 173.3 is: \[ P(M > 173.3) = 1 - 0.0107 = 0.9893 \] **In conclusion, there's about a 98.93% chance that the average of a sample of 20 values will be greater than 173.3.** --- ### Final summarized answers: - **Part (a):** 0.6966 - **Part (b):** 0.9893 Let me know if you'd like further clarification on any of these steps!